POJ 2442 Sequence

SequenceTime Limit: 6000 MS Memory Limit: 65536 KTotal Submissions: 5658 Accepted: 1722 Description Given m sequences, each contains n non-negative integer. now we may select one number from each sequence to form a sequence with m integers. it's clear that we may get n ^ m this kind of sequences. then we can calculate the sum of numbers in each sequence, and get n ^ m values. what we need is the small Est n sums. cocould you help us? Input The first line is an integer T, which shows the number of test cases, and then T test cases follow. the first line of each case contains two integers m, n (0 <m <= 100, 0 <n <= 2000 ). the following m lines indicate the m sequence respectively. no integer in the sequence is greater than 10000. output For each test case, print a line with the smallest n sums in increasing order, which is separ Ated by a space. sample Input 12 31 2 32 2 3 Sample Output 3 3 4 Source POJ Monthly, Guang Lin I wrote a heap, but it was tle. In fact, after understanding the principle, you don't need to heap. Using heap is not the minimum value. Just find the minimum value [cpp] # include <stdio. h> # include <string. h> # include <stdlib. h> int a [2100], B [2100], heap [1000000], data [110] [2100], Data [2100]; int n, m, pos [10000]; int cmp (const void * e, const void * f) {return (* (int *) e-* (int *) f );} int main () {void deal (int k); int I, j, s, t; scanf ("% d", & t); while (t --) {scanf ("% d", & m, & n); for (I = 1; I <= m; I ++) {for (j = 1; j <= n; j ++) {scanf ("% d", & Data [J-1]);} qsort (Data, n, sizeof (Data [0]), cmp); for (j = 1; j <= n; j ++) {data [I] [j] = Data [J-1] ;}} for (I = 1; I <= n; I ++) {a [I-1] = data [1] [I];} for (I = 2; I <= m; I ++) {deal (I) ;}for (I = 0; I <= n-1; I ++) {if (I = 0) {printf ("% d ", a [I]);} else {printf ("% d", a [I]);} printf ("\ n");} return 0 ;} void deal (int k) {int I, j, min, key; for (I = 1; I <= n; I ++) {pos [I] = 0 ;} for (I = 0; I <= n-1; I ++) {B [I] = data [k] [I + 1] ;}for (I = 1; I <= n; I ++) {heap [I] = a [I-1] + B [0];} for (I = 1; I <= n; I ++) {for (j = 1; j <= n; j ++) {if (j = 1) {min = heap [j]; key = j ;}else {if (heap [j] <min) {min = heap [j]; key = j ;}}} a [I-1] = heap [key]; heap [key] = heap [key]-B [pos [key] + B [pos [key] + 1]; pos [key] + = 1 ;}}